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Need help calculating angle with coefficient of static friction

  1. Feb 27, 2012 #1
    There's a box sitting on a plank. The coefficient of static friction is 0.33. I want to figure out at which angle the box will begin to slide if the plank is tilted.


    I know the factors that come into play are FN, Fg or Fgx, and Fs.



    So far I have

    Fnet= FN-Fs-Fg
    =FN- μs-sin∅
    =cos∅-0.33-sin∅

    This doesn't look right... can anyone help me out?
     
    Last edited: Feb 27, 2012
  2. jcsd
  3. Feb 27, 2012 #2

    jhae2.718

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    What are the conditions for impending motion using Newton's second law?

    I would suggest looking at the horizontal and vertical scalar equations of motion of the block and solving for [itex]\theta[/itex].

    I'm having a bit of difficulty following your work. Could you clarify your last expression? (It looks like [itex]F_{net} = \cos\theta-0.33-\sin\theta[/itex] to me.)
     
  4. Feb 27, 2012 #3
    Im assuming there's no acceleration, so Fnet= 0N.

    This is the symbol for theta: θ

    I used this symbol for "angle":∅

    """"I'm having a bit of difficulty following your work. Could you clarify your last expression? (It looks like F_{net} = \cos\theta-0.33-\sin\theta to me.) """"

    That's pretty much it! Except using angle to substitute for theta.

    I don't know how to get to the angle...

    I'm missing mass from the equation too right?
     
    Last edited: Feb 27, 2012
  5. Feb 27, 2012 #4

    jhae2.718

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    That's right. Impending motion, if you are unfamiliar with the term, is the point right before motion starts.

    There's one more condition dealing with static friction. What would that be? (Remember how static friction works; is there a maximum value? What is it?)

    The first place to start is by drawing a free body diagram of the situation. We know the block is tilted at some angle [itex]\theta[/itex] at which the box will just start to move, so what forces act on the block?

    Next, sum the forces in the [itex]x[/itex] and [itex]y[/itex] directions (I would suggest using a coordinate system aligned with the ramp.). From these two equations (plus an expression relating the frictional force and the normal force) you should be able to find [itex]\theta[/itex].
     
  6. Feb 27, 2012 #5
    0.33 ?

    Force of gravity in the x compenent, the normal force, and the force of friction.

    So you mean :

    x

    Fnet= Fgx
    =mg(sin∅)-Fs
    =mg(sin∅)-μs

    and

    y

    Fnet=FN
    0= mgcos∅

    So far so good?

    One more thing: The weight of the box is 10kg
     
  7. Feb 27, 2012 #6

    jhae2.718

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    [itex]\mu_s[/itex] is the coefficient of static friction, not the force of friction. So, what is the force of friction in this case?

    You're almost there!
     
  8. Feb 27, 2012 #7
    So.... since

    μs= Ff/FN
    0.33=Ff/mg
    Ff=0.33(10)(9.8)
    Ff=32.34N

    ?
     
  9. Feb 27, 2012 #8

    jhae2.718

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    Static friction is given by [itex]F_s \leqslant \mu_s N[/itex], so [itex]F_{s,max}=\mu_s N[/itex]. So, from this and the FBD we can get the friction force as a function of mass, gravitational acceleration, and the angle [itex]\theta[/itex].

    What is this force, and how can we substitute it to find [itex]\theta[/itex]?

    Hint: the final answer will be independent of mass and gravitational acceleration.
     
  10. Feb 27, 2012 #9
    You're saying

    Fsmax=m(μs)FN

    and i'm not sure what to subsitute the force with...
     
  11. Feb 27, 2012 #10
    so technically....

    if i know Fg and FN, I can figure out the angle?

    So

    tan∅= 32.34/9.8
    ∅=73°

    ?
     
  12. Feb 27, 2012 #11

    jhae2.718

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    Close, but not quite.

    [itex]F_{s,max} = \mu_sN = \mu_s mg\cos\theta[/itex].

    Then if [itex]ma = 0 = mg\sin\theta - \mu_s mg\cos\theta[/itex], what is [itex]\theta[/itex]?

    Sorry for the latency in response, just got out of class.
     
  13. Feb 27, 2012 #12
    No I really appreciate your help... getting my noggin working...

    Fsmax=musN
    =mus(mg)(costheta)

    0=mgsinθ-mμs(mg)cosθ

    that's the part where i'm confused... how do i isolate theta?
     
  14. Feb 27, 2012 #13

    jhae2.718

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    Maybe you could try moving [itex]\mu_s mg \cos\theta[/itex] to the other side and using trigonometry? Also, what can you do with the [itex]mg[/itex] term?
     
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